What is the value of $\dfrac{d}{dx}\cot(x)$ at $x=\dfrac{5\pi}{6}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{\sqrt{2}}{2}$ (Choice B) B $\dfrac{\sqrt{2}}{2}$ (Choice C) C $-4$ (Choice D) D $4$
Answer: Let's first find $\dfrac{d}{dx}\cot(x)$. Then, we can evaluate it at $x=\dfrac{5\pi}{6}$. Recall that the derivative of $\cot(x)$ is $-\dfrac{1}{\sin^2(x)}$, or $-\csc^2(x)$. Put another way, $\dfrac{d}{dx}[\cot(x)]=-\dfrac{1}{\sin^2(x)}=-\csc^2(x)$. [Is there a way to know this without memorizing?] Now let's plug in $x={\dfrac{5\pi}{6}}$ : $\begin{aligned} &\phantom{=}-\dfrac{1}{\sin^2\left({\dfrac{5\pi}{6}}\right)} \\\\ &=-\dfrac{1}{\left(\dfrac12\right)^2} \\\\ &=-\dfrac{1}{\dfrac14} \\\\ &=-4 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\cot(x)$ at $x=\dfrac{5\pi}{6}$ is $-4$.